# Question #ff200

Mar 4, 2017

Force on a proton in electric field $\vec{E}$ is given by
$\vec{F} = q \vec{E}$
Acceleration produced is $\vec{a} = \frac{q}{m} \vec{E}$
where $q \mathmr{and} m$ are the proton's charge and mass.

Treating orthogonal velocities independently.

Time taken by proton to travel horizontal distance $d$ to just exit out of electric field
$t = \frac{d}{v}$ ......(1)

Vertical component of velocity ${v}_{x t}$ attained in this time is found by the following kinematic expression
$v = u + a t$
${v}_{x t} = \frac{q}{m} | \vec{E} | \frac{d}{v}$
${v}_{x t} = \frac{q d}{m v} | \vec{E} |$ .....(2)

We know that angle of the resultant of two vectors $\vec{P} \mathmr{and} \vec{Q}$ having angle $\theta$ between them is given by the formula
$\tan \alpha = \frac{| \vec{Q} | \sin \theta}{| \vec{P} | + | \vec{Q} | \cos \theta}$

We get at the exit point
$\tan {15}^{\circ} = \frac{{v}_{x t} \sin {90}^{\circ}}{v + {v}_{x t} \cos {90}^{\circ}}$
$\implies \tan {15}^{\circ} = \frac{{v}_{x t}}{v}$
Using (2) we get
$\tan {15}^{\circ} = \frac{\frac{q d}{m v} | \vec{E} |}{v}$
$\implies \tan {15}^{\circ} = \frac{q d | \vec{E} |}{m {v}^{2}}$
$\implies | \vec{E} | = \frac{m {v}^{2} \tan {15}^{\circ}}{q d}$ .....(3)

Taking charge of proton $q = 1.602 \times {10}^{-} 19 C$,
mass of proton $m = 1.673 \times {10}^{-} 27 k g$
$| \vec{E} | = 932.7 V {m}^{-} 1$
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Alternate method after equation (1).

During time $t$ vertical deflection produced to meet the condition is
$\frac{\text{deflection}}{\frac{d}{2}} = \tan {15}^{\circ}$

as the line of exit velocity extended backwards meets the axis of velocity $\vec{v}$ at mid-point of $d$ i.e., at a distance $= \frac{d}{2}$

Setting up the kinematic equation for vertical deflection and noting that initial vertical velocity of proton is zero as it enters the uniform electric field region
$s = u t + \frac{1}{2} a {t}^{2}$
$\frac{d}{2} \tan {15}^{\circ} = \frac{1}{2} \left(\frac{q}{m} | \vec{E} |\right) {\left(\frac{d}{v}\right)}^{2}$
$d \tan {15}^{\circ} = \left(\frac{q}{m} | \vec{E} |\right) {\left(\frac{d}{v}\right)}^{2}$
$\implies | \vec{E} | = \frac{m}{q} d \tan {15}^{\circ} {\left(\frac{v}{d}\right)}^{2}$
$\implies | \vec{E} | = \frac{m {v}^{2} \tan {15}^{\circ}}{q d}$
which is same as equation (3)